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EMGT 5412Operations Management Science

Linear Programming: Introduction, Formulation and

Graphical Solution

Dincer KonurEngineering Management and Systems

Engineering 1

Outline• Definition of Linear Programming• Formulation of Linear Programming

– Surviving in an island– Extending a problem formulation – Wyndor Glass Co. Product Mix Problem

• Linear Programming Terminology• Graphical Solution to Linear Programming• Properties of Linear Programming

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What is Mathematical Programming?• Mathematical programming is selecting the best

option(s) from a set of alternatives mathematically– Minimizing/maximizing a function where your

alternatives are defined by functions• Many industries use mathematical programming

– Supply chain, logistics, and transportation– Health industry, energy industry, finance, airlines– Manufacturing industry, agriculture industry– Education, Military

• Operations Research raised with WWII

4

What is Linear Programming?• Linear Programming problem (LP) is a

mathematical programming problem where all of your functions are linear– A function is linear when

• The variables have power of 1 and the variables are not multiplied with each other

– linear– not linear– linear– not linear

– It is assumed that our variables are continuous (for now)

𝑓 (𝑥 )

𝑥

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How to formulate?• What do we need to formulate a linear

programming problem?– Have a problem!!– Know your problem

• Gather the relevant data• Know what each number means

– Pay attention to the class – Best way to learn is to do some examples….

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Example 1: Surviving in an island• Linear programming will save you!

– Suppose that you will be left in a deserted island

– You want to live as many days as you can to increase chance of rescue

– Fortunately (!), you can take a bag with you to the island

– What would you put into the bag?

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Example 1: Surviving in an island• The bag can carry at most 50 lbs• There are limited set of items you can carry

– Bread: You can survive for 2 days with 1 lb of bread– Steak: You can survive for 5 days with 1 lb of steak

(Memphis style grilled!)– Chicken: You can survive for 3 days with 1 lb of

chicken (southern style deep fried!)– Chocolate: You can survive for 6 days with 1 lb of

chocolate• How much of each item to take with you?

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Example 1: Formulation Steps• STEP 0: Know the problem and gather your data

– Your problem is to increase your chance of rescue by maximizing the number of days you survive

– You can have 1 bag which can carry 50 lbs at most– You can only put bread, steak, chicken and chocolate

into you bag– Each item enables you survive for a specific number of

days for each pound you takeBread: 2 days/lb Steak: 5 days/lbChicken: 3days/lb Chocolate: 6days/lb

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Example 1: Formulation Steps• STEP 1: Identify your decision variables

– Decision variables are the things you control– The amount of each item you will take with you

Warning: Always be careful with the metrics (try to use the same metrics)

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Example 1: Formulation Steps• STEP 2: Define your objective function and

objective– Your objective function is the measure of performance

as a result of your decisions– Your objective is what you want to do with your

objective function

– Recall that you want to maximize the number of days you will survive in the island

Objective

Performance measureNot a function though12

Example 1: Formulation Steps• STEP 2: We know our objective (maximize) and

our performance of measure (number of days)– Express the performance of measure as a function of

your decision variables to get the objective function• If you take lbs of bread, you will survive for days• If you take lbs of steak, you will survive for days• If you take lbs of chicken, you will survive for days• If you take lbs of chocolate, you will survive for days

– Then objective function in terms of decision variables:

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒13

Example 1: Formulation Steps• STEP 3: Define your restrictions (constraints)

– There may be some restrictions which limit what you can do (hence, they define set of your alternatives)

– You can carry 1 bag and it can carry 50 lbs at most

– You cannot get negative amounts!!!

Total amount you decide to carry

Limit on how much you can

carry

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Example 1: LP Formulation• Combine your objective, objective function, and

constraints

» This is your LP model» Note that everything is linear!

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜

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Example 1: Extending the LP• Suppose that you can also take cheese

– Cheese: You can survive for 4 days with 1 lb of cheese

– We have a new decision variable: Update LP

,


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Example 1: Extending the LP• Furthermore, suppose you have a budgetary limit

• 1 lb of bread costs you $3• 1 lb of steak costs you $6• 1 lb of chicken costs you $7• 1 lb of chocolate costs you $15• 1 lb of cheese costs you $8

– You can spend at most $150– Write the new restriction:

Total money you decide to spend

Limit on how much you can

spend

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Example 1: Extending the LP

,

– You cannot get more meat (chicken+steak) than bread

– For each lb of cheese, you need at least 2 lbs of bread


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Summary of Formulation Steps• Gather all of your data, know what they mean• Then

1. Identify your decision variables2. Identify your objective and objective function3. Identify your constraints

• Express your objective function and constraints in terms of your decision variables

• Steps 2 and 3 can change order

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• Wyndor Glass Co. produces high-quality glass products– The company decides to produce two new products

• A glass door with aluminum framing• A wood-framed glass window

– The company has 3 plants

Case 1: Wyndor Glass Co. Product

Plant Doors Windows Availability/week

1 1 hour 0 4 hours

2 0 2 hours 12 hours

3 3 hours 2 hours 18 hours

Production Time Used for Each Unit Produced

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Case 1: Wyndor Glass Co. Product• Unit profit for doors is $300• Unit profit for windows is $500

• What should be the product mix to maximize profits?– How many of each item to produce weekly?

• This is production rate, so it can be continuous, i.e., you can choose to produce 2.5 windows per week (this would be 10 windows per month assuming 4 weeks in a month)

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Case 1: Formulating the LP• Step 0: Gather the data (we have it all!)• Step 1: Decision variables

– D: the number of doors produced weekly– W: the number of windows produced weekly

• Step 2: Define the objective & objective functions– Objective: Maximization– Objective function: Profit=300D+500W

– It is the weekly profit

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Case 1: Formulating the LP• Step 3: Define the constraints

– Plant 1 capacity: D ≤ 4– Plant 2 capacity: 2W ≤ 12– Plant 3 capacity: 3D+2W ≤ 18

• Combine what you have!Maximize 300D+500W

subject to D ≤ 4 2W ≤ 12

3D+2W ≤ 18 D≥0, W ≥0

Do not forget the non-negativity

constraints 23




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LP Terminology• Decision Variables: things we control• Objective function: measure of performance• Nonnegativity constraints• Functional constraints: restrictions we have• Parameters: constants we use in the objective function

and constraint definitions• Solution: any choice of values for the decision variables

– Feasible solution is one that satisfies the constraints– Optimal solution is the best feasible solution

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Further Study• Read 2.2:

– Formulate the Profit & Gambit Co. Advertising-Mix problem for practice,

– Practice problems:2.5, 2.7, 2.8, 2.12, 2.22, 2.24, 2.27

– Do not worry about the spreadsheet questions• Practice case:

– Case 2-3 Staffing a Call Center (try to formulate the problem)

• Solutions to Practice problems and cases are posted under Practice Files folder for each chapter…

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Challenge• Below is a question I was asked during on-site

interview with Amazon in Seattle, WA– Suppose you have a set of integer numbers– Formulate an LP that finds the minimum number in this

set• Note: your decision variables should be continuous

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Graphically Solving Simple LPs• So far we have discussed how to formulate LPs• How about solving them?• Simple LPs can be solved graphically

– Simple LP: 2 decision variables!1. Graph the constraints to see which points satisfy all

constraints (i.e., graph the feasible points)2. Graph iso-lines of the objective function (objective

function lines)3. Use iso-lines to find the optimal solution(s)

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Example 2: Product Mix Problem• Maximize your profit

– Decide how many to produce product types 1,2– To be produced, each product requires different

amount of 2 resources per unit– Each resource is limited

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Example 2: LP formulation• You can formulate the following LP model

• type 1 produced (units)• (units)

Resource A limit

Resource B limit

Non-negativity

Total profit

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Example 2: Graphical Solution• First, draw your

decision variables as the axes of your graph

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Example 2: Graphical Solution• Start with the easy

constraints– Non-negativity

constraints!– Assume equality of

the constraint ( or )

– Then, find the region defined by the constraint

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Example 2: Graphical Solution• Resource A limit

constraint– Draw – Find the region

where – Hint: take a point in each of

one of the regions and see if this point satisfies the constraint. The region including the point satisfying your constraint is the region defined by your constraint (you can take the origin as your point)

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Example 2: Graphical Solution• Resource B limit

constraint– Draw 2 – Find the region

where 2

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Example 2: Graphical Solution• Define the feasible

region– It is the region

defined by the intersection of all of your constraints

– It includes all of the feasible solutions to your LP

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Example 2: Graphical Solution

Improvement Direction

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Example 2: Graphical Solution• Optimum solution

will be one of the corner solutions!!1. Graph the

constraints2. Define your feasible

region3. Evaluate the corner

points (or draw iso-lines until you leave feasible region)

4. Choose the best corner point

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Graphical Solution to Case 1• Recall the Wyndor Glass Co. Product Mix

problem– D: number of doors, W: number of windows

Maximize 300D+500W subject to D ≤ 4

2W ≤ 12 3D+2W ≤ 18 D≥0, W ≥0

Plant 1 availability

Total profit

Plant 2 availabilityPlant 3 availability

Non-negativity

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Graphical Solution to Case 1• STEP 1&2: Graph the constraints and find feasible region

0 2 4 6 8

8

6

4

10

2

Feasible

region

Production rate for doorsD

W

2 W =12

D = 4

3 D + 2 W = 18

Production rate for windows

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Graphical Solution to Case 1• STEP 3: Draw iso-lines to find the optimal solution

0 2 4 6 8

8

6

4

2

Production rate

for windows

Production rate for doors

Feasible

region

(2, 6)

Optimal solution

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W

D

P = 3600 = 300D + 500W

P = 3000 = 300D + 500W

P = 1500 = 300D + 500W

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Summary of the Graphical Method• Draw the constraint boundary line for each constraint. Use the origin (or any

point not on the line) to determine which side of the line is permitted by the constraint.

• Find the feasible region by determining where all constraints are satisfied simultaneously.

• Determine the slope of one objective function line (iso-line). All other objective function lines will have the same slope.

• Move a straight edge with this slope through the feasible region in the direction of improving values of the objective function. Stop at the last instant that the straight edge still passes through a point in the feasible region. This line given by the straight edge is the optimal objective function line.

• A feasible point on the optimal objective function line is an optimal solution.

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Further Study• Graphical Method:

– Read 2.4 – Practice Problems:

• 2.9.h, 2.13.a, 2.19.e, 2.26.a • See the video if you want someone else talk about graphical

method: http://www.youtube.com/watch?v=gz6_uXyK9yw – Practice drawing lines!

• Online tools are available

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http://www.youtube.com/watch?v=gz6_uXyK9yw

http://www.youtube.com/watch?v=gz6_uXyK9yw




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Properties of LPs• An LP problem can have

– Infeasibility• No feasible solutions!

– Unique optimal solution• Only one feasible solution is optimum

– Multiple optimal solutions (alternative optima)• More than one feasible solutions that are optimum

– Unboundedness• There is always a better feasible solution

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Infeasibility• Consider the following LP problem

– Practice: draw the feasible region

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Infeasibility• The feasible

region is empty• That is, no

solution satisfies the constraints (i.e., there is no feasible solution)– So, no optimal

solution exists47

Unique Optimal Solution• Consider the following LP problem (Example 2)

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Unique Optimal Solution• The feasible region

is not empty• All of the points in

the feasible region are feasible solutions

• But we have a single optimal solution

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Alternative Optima• Consider the following LP problem

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Alternative Optima• The feasible region

is not empty– There are many

feasible solutions– Also, there are

many optimal solutions • but still there are

corner points that are also optimum!

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Challenge• Does an LP have to have at least two corner

solutions that are optimum in case of alternative optima? (I asked this in the Ph.D. qualifying exam)– No!– Proof?– Think about a counter example

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Unboundedness• Consider the following LP problem

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Unboundedness• The feasible region

is unbounded and– You can increase

the objective function value as much as you can while you are still in the feasible region

– Unbounded feasible region does not mean unboundedness of LP

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Further Properties of LPs• Given that the constraints do not change

– Maximizing 3A+4B = Minimizing -3A-4B• Maximization is opposite of minimization

– Maximizing 3A+4B = Maximizing 100+3A+4B• Adding a constant to the objective function does not change

the optimum solution, you can just ignore it for optimization– Maximizing 3A+4B = Maximizing 15A+20B

• Multiplying all of the coefficients of the objective function with the same positive constant does not change the optimization

• On another note, you can multiply both sides of a constraint with the same non-zero constant and it will not change the feasible region, i.e., A+B<= 5 defines the same region with 2A+2B<=10 55

Further Properties of LPs• Divisibility Assumption of Linear

Programming: – Decision variables in a linear programming model are

allowed to have any values, including fractional values, that satisfy the functional and nonnegativity constraints. Thus, these variables are not restricted to just integer values.

– If violated, you do not have an LP

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Further Properties of LPs• LP models are polynomially solvable

– That is, they are relatively easy to solve– Simplex Method is a very popular method to solve LP

problems• If an optimum solution exists to an LP problem, then there

exists an optimum corner solution (this is true for any LP)– Start with a corner solution (extreme point), move to a better

corner solution (there are finite corner solutions)– Repeat this until you cannot find a better corner solution– http://en.wikipedia.org/wiki/Simplex_algorithm

– Karmarkar’s algorithm and Ellipsoid Method are other ways to solve LP models (wiki them to have a look)

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http://en.wikipedia.org/wiki/Simplex_algorithm

http://en.wikipedia.org/wiki/Simplex_algorithm

Software for solving LPs• There are many software for solving LPs• They mostly use Simplex type of algorithms

– We will learn how to solve simple LPs in Excel– Excel solver uses Simplex Algorithm

• Other software– CPLEX (one of the best solvers for LPs)– Matlab has a function to solve LPs (linprog)– Optimization software can solve LPs

• GAMS, Xpress, Lindo, Lingo, AMPLE

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Next time….• Spreadsheet modeling

– Using Excel solver to solve LPs• Applications

– Different cases with different LP formulations…• Sensitivity Analysis (briefly)• Preview Chapter 3

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